Newton's laws of motion | Second law

Eleven Standard >> Newton's laws of motion | Second law

Understanding Impulse and Impulsive Force: A Mathematical Dive into Newton’s Second Law

Newton’s Second Law of Motion is often summarized as F = ma (force equals mass times acceleration). However, its more general form relates force to the rate of change of momentum:

\(F=\frac{dp}{dt}\)

Here, momentum (p) is the product of mass (m) and velocity (v): p = mv. This equation implies that force is directly tied to how quickly an object’s momentum changes over time.

Momentum and the Second Law

When mass is constant, the law simplifies to F = ma, as acceleration (a) is the derivative of velocity. However, the momentum form ( \(F=\frac{dp}{dt}\)) is crucial for understanding scenarios involving varying mass or velocity, such as rocket motion or collisions.

Impulse: The Change in Momentum

Impulse (J) quantifies the total effect of a force acting over time. Mathematically, it’s the integral of force over a time interval (Δt):

\(J=\int_{t_1}^{t_2} F dt\)

From Newton’s Second Law, this integral equals the change in momentum (Δp):

J = Δp = p_final − p_initial

For constant force, impulse simplifies to: J = F_avg × Δt

Key Insight:

• Units: Impulse is measured in Newton-seconds (Ns) or kg·m/s, matching momentum’s units.
• Graphical Representation: The area under a Force-Time graph equals impulse.

Impulsive Force: Large Forces in Short Times

An impulsive force acts over an extremely short duration (Δt → 0), producing a significant change in momentum. Examples include hammer strikes, car crashes, or a cricket bat hitting a ball. Using the impulse equation:

F_avg = Δp / Δt

If Δt is tiny, F_avg becomes very large even for modest momentum changes.

Example Calculation:

A cricket ball (m = 0.15 kg) accelerates from 0 to 20 m/s in 0.01 s:

F_avg = (0.15 × 20) / 0.01 = 3 / 0.01 = 300 N

Real-World Applications

• Sports: Cushioned gloves increase impact time, reducing force (e.g., in boxing).
• Automotive: Airbags extend collision time, minimizing injury.
• Engineering: Shock absorbers dissipate impulsive forces in machinery.

Mathematical Problems and Solutions

Problem 1:

A 5 kg object experiences a force of 10 N for 3 s. Find its impulse and final velocity (if initial velocity is 2 m/s).

Solution:

Impulse: J = F × Δt = 10 × 3 = 30 Ns

Δp = 30 = m(v_f − v_i)

⇒ v_f = (30 / 5) + 2 = 8 m/s